The radius of hydrogen atom in its ground state is `5.3 xx 10^(-11)m`. After collision with an electron it is found to have a radius of `21.2 xx 10^(-11)m`. What is the principle quantum number of `n` of the final state of the atom ?

Options

A. `n=4`
B. `n=2`
C. `n=16`
D. `n=3`

Answer

B

Explanation

`r_(n) prop n^(2) or n prop sqrt(r_(n))`. If `r_(i)` is the intial orbital radius and `r_(f)` is the final orbital radius, then the principal quantum number of the final state is
`n=sqrt((r_(f))/(r_(i)))=((21.2xx10^(-11))/(5.3xx10^(-1)))^(1//2)=2`